Editing Fubini's theorem (section)

==Counterexamples==

The following examples show how Fubini's theorem and Tonelli's theorem can fail if any of their hypotheses are omitted.

===Failure of Tonelli's theorem for non σ-finite spaces===

Suppose that ''X'' is the unit interval with the Lebesgue measurable sets and Lebesgue measure, and ''Y'' is the unit interval with all the subsets measurable and the [[counting measure]], so that ''Y'' is not σ-finite. If ''f'' is the characteristic function of the diagonal of ''X''×''Y'', then integrating ''f'' along ''X'' gives the 0 function on ''Y'', but integrating ''f'' along ''Y'' gives the function 1 on ''X''. So, the two iterated integrals are different. This shows that Tonelli's theorem can fail for spaces that are not σ-finite no matter which product measure is chosen. The measures are both [[decomposable measure|decomposable]], showing that Tonelli's theorem fails for decomposable measures (which are slightly more general than σ-finite measures).

===Failure of Fubini's theorem for non-maximal product measures===

Fubini's theorem holds for spaces even if they are not assumed to be σ-finite provided one uses the maximal product measure. In the example above, for the maximal product measure, the diagonal has infinite measure so the double integral of |''f''| is infinite, and Fubini's theorem holds vacuously.
However, if we give ''X''×''Y'' the product measure such that the measure of a set is the sum of the Lebesgue measures of its horizontal sections, then the double integral of |''f''| is zero, but the two iterated integrals still have different values. This gives an example of a product measure where Fubini's theorem fails.

This gives an example of two different product measures on the same product of two measure spaces. For products of two σ-finite measure spaces, there is only one product measure.

===Failure of Tonelli's theorem for non-measurable functions===

Suppose that ''X'' is the first uncountable ordinal, with the finite measure where the measurable sets are either countable (with measure 0) or the sets of countable complement (with measure 1). The (non-measurable) subset ''E'' of ''X''×''X'' given by pairs (''x'' , ''y'') with ''x''<''y'' is countable on every horizontal line and has countable complement on every vertical line. If ''f'' is the characteristic function of ''E'' then the two iterated integrals of ''f'' are defined and have different values 1 and 0. The function ''f'' is not measurable. This shows that Tonelli's theorem can fail for non-measurable functions.

===Failure of Fubini's theorem for non-measurable functions===

A variation of the example above shows that Fubini's theorem can fail for non-measurable functions even if |''f''| is integrable and both repeated integrals are well defined: if we take ''f'' to be 1 on ''E'' and –1 on the complement of ''E'', then |''f''| is integrable on the product with integral 1, and both repeated integrals are well defined, but have different values 1 and –1.

Assuming the continuum hypothesis, one can identify ''X'' with the unit interval ''I'', so there is a bounded non-negative function on ''I''×''I'' whose two iterated integrals (using Lebesgue measure) are both defined but unequal. This example was found by {{harvs|txt|first=Wacław |last=Sierpiński |authorlink=Wacław Sierpiński|year=1920}}.<ref>{{citation |first=Wacław |last=Sierpiński |authorlink=Wacław Sierpiński |year=1920 |title=Sur un problème concernant les ensembles mesurables superficiellement |journal=[[Fundamenta Mathematicae]] |volume=1 |issue=1 |pages=112–115 |doi=10.4064/fm-1-1-112-115 |url=https://eudml.org/doc/212592 |doi-access=free }}</ref>
The stronger versions of Fubini's theorem on a product of two unit intervals with Lebesgue measure, where the function is no longer assumed to be measurable but merely that the two iterated integrals are well defined and exist, are independent of the standard [[Zermelo–Fraenkel axioms]] of [[set theory]].  The continuum hypothesis and [[Martin's axiom]] both imply that there exists a function on the unit square whose iterated integrals are not equal, while {{harvs|txt|first=Harvey |last=Friedman | authorlink=Harvey Friedman (mathematician)|year=1980}} showed that it is consistent with ZFC that a strong Fubini-type theorem for [0,1] does hold, and whenever the two iterated integrals exist they are equal.<ref>{{citation |first=Harvey |last=Friedman | authorlink=Harvey Friedman (mathematician)| year=1980 |title=A Consistent Fubini-Tonelli Theorem for Nonmeasurable Functions |journal=[[Illinois Journal of Mathematics]] |volume=24 |issue=3 |pages=390–395 |doi=10.1215/ijm/1256047607 |mr=573474|url=http://projecteuclid.org/euclid.ijm/1256047607 |doi-access=free }}</ref> See [[List of statements undecidable in ZFC]].

===Failure of Fubini's theorem for non-integrable functions===
Fubini's theorem tells us that (for measurable functions on a product of σ-finite measure spaces) if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect to ''x'' and then with respect to ''y'', we get the same result as if we integrate first with respect to ''y'' and then with respect to ''x''. The assumption that the integral of the absolute value is finite is "[[Lebesgue integral|Lebesgue integrability]]", and without it the two repeated integrals can have different values.

A simple example to show that the repeated integrals can be different in general is to take the two measure spaces to be the positive integers, and to take the function ''f''(''x'',''y'') to be 1 if ''x'' = ''y'', −1 if ''x''&nbsp;=&nbsp;''y''&nbsp;+&nbsp;1, and 0 otherwise. Then the two repeated integrals have different values 0 and 1.

Another example is as follows for the function
<math display="block">\frac{x^2-y^2}{(x^2+y^2)^2} = -\frac{\partial^2}{\partial x\,\partial y} \arctan(y/x).</math>
The [[double integral|iterated integral]]s

<math display="block">\int_{x=0}^1\left(\int_{y=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\text{d}y\right)\,\text{d}x = \frac{\pi}{4}</math>
and
<math display="block">\int_{y=0}^1\left(\int_{x=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\text{d}x\right)\,\text{d}y=-\frac{\pi}{4}</math>
have different values. The corresponding double integral does not [[absolute convergence|converge absolutely]] (in other words the integral of the [[absolute value]] is not finite):
<math display="block">\int_0^1\int_0^1 \left|\frac{x^2-y^2}{\left(x^2 + y^2\right)^2}\right|\,\text{d}y\,\text{d}x=\infty.</math>